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Calculate zeros of Riemann zeta function with 500 decimal digits accuracy?

Posted 6 years ago

Dear Friends Recently I have published an article related the imaginary parts of the zeros of the Riemann zeta function . http://ijmcs.future-in-tech.net/13.1/R-Gunes.pdf Here , we have found a new formula which is very simple formula ( equation 3.3 in page 3).


Then, we have decided to find the imaginary parts of the zeros of the Riemann zeta function by this equation 6.1 in page 7.

Now, can you arrange a computer calculation program which will allow us will see the calculations with 500 decimal digits accuracy. Thus if we use the calculation results of Prof Odlyzko about the zeros of the Riemann zeta function in this calculation program, can we see that the Riemann zeta function will be equal zero with 500 or 1000 decimal digits accuracy or not ?.

Shortly we can say here, without using Riemann-Siegel Function Z(t) , we can find the zeros of the Riemann zeta function.

Maybe, we can make just 6 calculations. These 6 calculations will be enough to see the sensitivity of this new formula . enter image description here

Thank you again for your kind support. I am always ready here to reply your all questions.

POSTED BY: Bahattin GUNES
5 Replies

Bahattin,

I have not read your paper but I assume that you are asking how to get 500 digits of accuracy in your calculation. If this is correct, you should use WorkingPrecision and decimal numbers with 500 digit accuracy. For example, 0.1 has only 16 digits of accuracy but

0.1`500  

has 500, 1/2 is an exact number (while

0.5`500

has 500 digits of accuracy)

In[1]:= NIntegrate[Cos[0.1`500 Log[x]]/Sqrt[x], {x, 0, 100 + 1/2}, 
 WorkingPrecision -> 500]

Out[1]= 18.\
9832766381160522781389347000142510618943050267644478031788925490903048\
7088013603603244391365065713045986514405142264574345939164757619909592\
1457936292194070591365911489826615498548791270507814145953297444844607\
4361220007631734791464226409028830961397455341828481486671422944178478\
6562735625536587471599195629078713825602237936923987911506096841820494\
9623120521855209468800917396895715542716090191257440339460451160356487\
2339726206109965665579270766406867302307529026669144918909126600514810\
01679179

I hope this is what you are looking for.

Regards

POSTED BY: Neil Singer
Posted 6 years ago

Dear Neil

Thank you for your comment. Please you can see my second letter above what I am looking exactly. Could you help me , please...

Sincerly, Bahattin

POSTED BY: Bahattin GUNES

I'm almost certain that Mathematica can do real calculations with that sort of precision but unfortunately I don't yet know how to set the precision for such calculations.

I am more familiar with Number Theory Library (NTL) which does arbitrary precision arithmetic with both integers and reals. NTL does not do symbolic math though, so for instance you.d have to do the integrals on the RHSs by hand (or a symbolic math program) and then code it in C++, calling NTL where needed. FWIW, NTL is GPL3.

There's also Axiom, for which there is a TeXmacs plugin. Axiom documentation is Huge. Finding how to set the precision may be difficult.

Best of luck,,and contact me here if you want me to elaborate on any of this. Regards, Graham.

POSTED BY: Graham Cottrell
Posted 6 years ago

Dear Nail , Dear Graham Dear Friend Could you help me ? I want to see the following calculation results with the ´500´ decimal digits ( I have no any idea about computer programming) Thank you very much for your this help.

$ s_1=0.5+it_1$ and $ s_2=0.5+it_2$
See $ t_1$ and $ t_2$ on below as with the ´500´ decimal digits .

$ N_1=10^{14}$ , $ N_2=10^{16}$ and $ N_3=10^{18}$ enter image description here

Or if you consider, I can give the real functions instead of the above - complex function).

These total 6 calculations will be enough for me. If you have any question, please do not hesitate.

$ t_1$=14.134725141734693790457251983562470270784257115699243175685567460149 9634298092567649490103931715610127792029715487974367661426914698822545 8250536323944713778041338123720597054962195586586020055556672583601077 3700205410982661507542780517442591306254481978651072304938725629738321 5774203952157256748093321400349904680343462673144209203773854871413783 1735639699536542811307968053149168852906782082298049264338666734623320 0787587617920056048680543568014444246510655975686659032286865105448594 4432062407272703209427452221304874872092412385141835146054279015244783

$ t_2$=236.52422966581620580247550795566297868952949521218912370091896098781 9150384292332826261444604065174015827315678371311908125834666026514272 3427442710559412695531311686041161208386105258022537008648905356952178 6290474725602526826563480477997212116546341401730768688207342088012792 6644985268669489487216328558298806627153861486608112087260596055996397 4174141107331932566901602812030379738385137490926579562946886420201539 1274403467040835271042467991402807619705981381817538187168651059206275 5648099062015611059642060030641789502508561511728245696694558507994532

POSTED BY: Bahattin GUNES

Dear Bahattin,

Sorry to have taken so long to get back to you. I think I will have a bit more time for this from here on. I've noticed a few oddities and I'm afraid that I have some bad news.

The bad news first: the sum, counting up to N, for the large values of N that you want, is going to take too long for a single machine, especially since it involves a reciprocal and a power. Consider this: if incrementing n takes 1 clock; testing it against N another clock, raising n to the s'th another two clocks, taking the reciprocal another clock and adding the result to an accumulator yet another clock - that's six clocks per iteration, assuming we're using ordinary ints and doubles. A fast machine these days doesn't run much faster than 4GHz, so our loop takes 1.5e-9 seconds per iteration. For your lowest N, 1.0e14, it would take 1.5e5 seconds, or about 2500 minutes, or 41 hours 40 minutes. You can expect the times to be hundreds of times worse for 500-digit precision. It will also be hundreds of times worse for 1.0e16, ditto for N==1.0e18:-(

Cloud computing might just save you here, and I understand that Mathematica can do some cloudy stuff.

I noticed that you originally had a graphic containing two equations (real & imaginary parts?) in your first post here; now it's changed to rendered postscript or something. Anyhow, as I recall it there was an integral expression on the right. Now there's a dx on the right which doesn't seem to refer to anything.

Regards, Graham.

POSTED BY: Graham Cottrell
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